Question #10e47

2 Answers
Jan 18, 2018

Toque applied = 6 N.m & power is about120 watt

Explanation:

Here,by applying the torque angular retardation was done to stop its motion(considering no skidding was there).

so, angular retardation= alpha = change in angular velocity i.e omega/t = 40/10 i.e 4 radian/s^2

Now, torque = Ialpha (where, I is the moment of inertia)

moment of inertia of a cylinder passing through its axis is (Mr^2)/2 =1.5 Kg.m^2

so. tau = 1.5*4 N.m i,e 6 N.m

so,angular power required = (tau* theta)/t i.e about 120 watt(tau*theta= work done,dividing by time required for this angular displacement of theta by the time t in which it was brought to rest due to application of this torque) ( theta is angular displacement)( theta = (omega)^2/(2 alpha)=(40^2)/(2*4)=200 radian)

Jan 18, 2018

The torque is =6Nm. The power is =240W

Explanation:

The torque and the angular acceleration are related

"Torque (Nm)"="Moment of inertia (kgm^2)"xx "angular acceleration (rads^-2)"

tau=Ixxalpha

The moment of inertia of the solid cylinder is

I=(mr^2)/2

The mass of the cylinder is m=3kg

The radius of the cylinder is r=1.0m

The moment of inertia is I=(3xx1^2)/2=1.5kgm^2

The initial angular velocity is omega_0=40rads^-1

The final angular velocity is omega_1=0rads^-1

The time is t=10s

Apply the equation (to find the angular acceleration (alpha))

omega_1=omega_0+alphat

alpha=(omega_1-omega_0)/t=(0-40)/10=-4rads^-2

The torque to be applied is

tau=Ialpha=1.5xx4=6Nm

"Power (W)"= "torque (Nm)"xx "angular velocity (rads^-1)"

P=tauxx omega_0

The power is P=6xx40=240W